What is the solution for y^{''}+9y=sin(3x) ?

The solution for y^{''}+9y=sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)-(xcos(3x))/6

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y^{''}+9y=sin(3x)

y^{^{''}} + 9 y = sin (3 x)

y = c_{1} cos (3 x) + c_{2} sin (3 x) - \frac{x cos ( 3 x )}{6}

Solution steps

y^{''} (x) + 9 y = sin (3 x)

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) - \frac{x cos ( 3 x )}{6}

y = c_{1} cos (3 x) + c_{2} sin (3 x) - \frac{x cos ( 3 x )}{6}

(x, y) \to (5, - 2) lim (x^{5} + 4 x^{3} y - 5 x y^{2})

\int \frac{a x}{x ^{2} - p x} d x

x \to 3 - lim (\frac{x + 4}{x - 3})

\frac{7}{x ^{2}}, 1

\int 4 x^{\frac{1}{2}} d x

What is the solution for y^{''}+9y=sin(3x) ?
The solution for y^{''}+9y=sin(3x) is y=c_{1}cos(3x)+c_{2}sin(3x)-(xcos(3x))/6